Đáp án đúng: D
6,64.
Chú ý : $\displaystyle 2ROH\to R\text{OR}+{{H}_{2}}O$
Ta có :$\displaystyle {{n}_{{{H}_{2}}}}=0,15\,\,\,\,\,\,\,\,\to {{n}_{ancol}}=0,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to {{n}_{{{H}_{2}}O}}=0,15$
$\displaystyle \xrightarrow{BTKL}11=m+0,15.18\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{H=80{}^{o}\!\!\diagup\!\!{}_{o}\;}m=8,3.80{}^{o}\!\!\diagup\!\!{}_{o}\;=6,64$